Let $(X,\mu)$ be a measure space and let $f_j$,$1\leq j\leq m$ be measurable functions on X.
For $0<p<\infty.$Show that $$\bigg\lVert\max_{1\leq j\leq m}|f_j|\bigg\rVert_{L^{p,\infty}}^{p}\leq\sum_{j=1}^{m}\lVert f_j\rVert_{L^{p,\infty}}^p$$
Here is a detail proof by myself :
Now we use the notation $\omega_{|f|}(\alpha)=\mu\bigg\{x\in X:|f(x)|>\alpha\bigg\}$
For each $\alpha>0$,we have
$$\bigg\{x\in X :\max_{1\le j\le m}|f_j(x)|>\alpha \bigg\}\subseteq\bigcup_{j=1}^{m}\bigg\{x\in X:|f_j(x)|>\alpha \bigg\}$$
Whence, for each $\alpha>0,$one has
\begin{align} \alpha^{p}\omega_{\max |f_j|}(\alpha)&\leq\alpha^p\sum_{j=1}^{m}\omega_{|f_j|}(\alpha)\\ &=\sum_{j=1}^{m}\alpha^{p}\omega_{|f_j|}(\alpha)\\ &\leq\sum_{j=1}^{m}\sup\bigg\{\alpha^{p}\omega_{|f_j|}(\alpha) : \alpha>0\bigg\}\\ &=\sum_{j=1}^{m}\bigg(\sup\bigg\{\alpha\bigg(\omega_{|f_j|}(\alpha)\bigg)^{\frac{1}{p}}: \alpha>0\bigg\}\bigg)^{p}\\ &=\sum_{j=1}^{m}\lVert f_j\rVert_{L^{p,\infty}}^p \end{align}
Henceforth, our proof is complete when we take sup of all $\alpha>0$ of the both sides.
Is there anyone has the time to check my working for validity ? any advice and suggestion will appreciate. Thanks a lot.
It looks good, the rest issue is whether we are legitimate to write $\sup\{\alpha^{p}\omega_{|f_{j}|}(\alpha): \alpha>0\}$ to be $\left(\sup\left\{\alpha\left(\omega_{|f_{j}|}(\alpha)\right)^{1/p}\right\}\right)^{p}$. Indeed, this is taken as granted, but more precisely, one should consider if any $\|f_{j}\|_{L^{p,\infty}}=\infty$, there is nothing to prove. If it were less than $\infty$, then all the elements in the supremum, $\omega_{|f_{j}|}(\alpha)$ cannot be $\infty$, and the change of suprema is justified.
The issue is, what does it mean by $\infty^{1/p}$. Of course, the $p$ here is not zero, so we are pretty safe to say it is still $\infty$. However, if you define the positive power of $\infty$ to be $\infty$, one should review whether our theory of real analysis satisfies the consistency. And in fact, luckily, it does.